1) Without consulting a table of values, predict the relative trend of entropy per mole for
molecular oxygen, ozone, and monatomic oxygen gases under identical conditions.
Correct Answer:
O < O2 < O3
According to the third law of thermodynamics, entropy increases with increasing molecular
vibrations [which cease at absolute zero]. The more bonds a molecule has, the more ways the
molecule can vibrate. Since ozone has more bonds than the other two, it has the highest entropy,
followed by molecular oxygen—which has one bond—and then monatomic oxygen with the
lowest entropy.
2) For each of the following pairs, choose the substance with the higher entropy per mole at a
given temperature:
a) O2(g) at 5 atm or O2 (g) at 0.5 atm,
b) Br2 (l) or Br2 (g)
Correct Answer:
O2 (g) at 0.5 atm,
Br2 (g)
3) Calculate G° for the reaction below at 25°C.
2SO2 (g) +
O2 (g) -->
2SO3(g)
Substance
SO2 (g)
Hf°(kJ/mol) S°(J/mol·K)
-297
249
O2 (g)
0
205
SO3(g)
-395
256
Correct Answer:
-139 kJ You should calculate G for each substance first, multiply by the number of moles
present in the balanced chemical equation, and then calculate G for the reaction (Grxn =
Gproducts - Greactants).
3) Calculate Keq at 298 K for the reaction:
H2 (g) +
I2 (g) =
2HI(g).
G°
0
19.37kJ
1.30kJ
Corrrect Answer:
G° = 2(1.30 kJ) - 1(19.37 kJ) = -16.77 kJ
G° =-RT lnK => K=exp(-G° /RT) = EXP(16770/(8.314*298))  K= 8.70 x 102
4) Given the following standard molar entropies, calculate S in J/K for the reaction:\
S° J/mol·K
C(s) + O2 (g) =>
CO2 (g)
5.74
213.6
Correct Answer: 2.9
205.0
You need to subtract the reactants from the products..
5) Under what conditions would a reaction that is spontaneous at low temperatures and
becomes nonspontaneous at a higher temperature? (Assume that H and S do not change with
temperature.)
Correct Answer:
H < 0, S < 0
6) The reaction is never spontaneous.
Use the following data to calculate G° in kJ at 155°C for the reaction:
2O3(g) --> 3O2 (g)
Hf° kJ/mol
S° J/mol·K
142.7
238.8
Correct Answer:
0
205.0
 H = -285.4kJ ; S= 137.4J/K
G= H-T S = -344.207 kJ
7) The molar heat of vaporization for ethanol is 37.4 kJ/mol. If the boiling point of ethanol is
78.0°C, calculate S in J/K for the vaporization of 0.500 mol ethanol.
S = 37.4/(273+78) divided by 2 = 53.3 J/K
8) Calculate G° in kJ for the following reaction at 25°C if the value of Keq is 6.1 x 10-58.
3O2 (g) --> 2O3(g)
G = -RT lnK
G = =8.314*298*LN(6.1E-58)/1000 = -326.4 kJ